前言

本文隶属于专栏《LeetCode 刷题汇总》,该专栏为笔者原创,引用请注明来源,不足和错误之处请在评论区帮忙指出,谢谢!

本专栏目录结构请见LeetCode 刷题汇总

正文

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41. 缺失的第一个正数

题解

官方题解

3 次遍历,小于0都取大,所有取绝对值,小值取反,返回第一个大于0

class Solution {
    public int firstMissingPositive(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] <= 0) {
                nums[i] = n + 1;
            }
        }
        for (int i = 0; i < n; ++i) {
            int num = Math.abs(nums[i]);
            if (num <= n) {
                nums[num - 1] = -Math.abs(nums[num - 1]);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] > 0) {
                return i + 1;
            }
        }
        return n + 1;
    }
}

42. 接雨水

题解

官方题解

动态规划 + 3 次遍历

class Solution {
    public int trap(int[] height) {
        if(height.length == 0) return 0;
        int n = height.length;
        int[] leftMax = new int[n], rightMax = new int[n];
        leftMax[0] = height[0];
        for(int i = 1; i < n; i++){
            leftMax[i] = Math.max(leftMax[i - 1], height[i]);
        }
        rightMax[n - 1] = height[n - 1];
        for(int i = n - 2; i >= 0; i--){
            rightMax[i] = Math.max(rightMax[i + 1], height[i]);
        }
        int res = 0;
        for(int i = 0; i < n; i++){
            res += Math.min(leftMax[i], rightMax[i]) - height[i];
        }
        return res;
    }
}

单调栈

class Solution {
    public int trap(int[] height) {
        if(height.length == 0) return 0;
        int n = height.length, res = 0;
        ArrayDeque<Integer> stack = new ArrayDeque<>(n);
        for(int i = 0; i < n; i++){
            while(!stack.isEmpty() && height[i] > height[stack.peek()]){
                int top = stack.pop();
                if(stack.isEmpty()){
                    break;
                }
                int left = stack.peek();
                int width = i - left - 1;
                int h = Math.min(height[i], height[left]) - height[top];
                res += width * h;
            }
            stack.push(i);
        }
        return res;
    }
}

双指针 leftMax > rightMax

class Solution {
    public int trap(int[] height) {
        if(height.length == 0) return 0;
        int n = height.length, left = 0, leftMax = 0, right = n - 1, rightMax = 0, res = 0;
        while(left < right){
            leftMax = Math.max(leftMax, height[left]);
            rightMax = Math.max(rightMax, height[right]);
            if(leftMax > rightMax){
                res += rightMax - height[right];
                right--;
            }else{
                res += leftMax - height[left];
                left++;
            }
        }
        return res;
    }
}

双指针 height[left] > height[right]

class Solution {
    public int trap(int[] height) {
        if(height.length == 0) return 0;
        int n = height.length, left = 0, leftMax = 0, right = n - 1, rightMax = 0, res = 0;
        while(left < right){
            leftMax = Math.max(leftMax, height[left]);
            rightMax = Math.max(rightMax, height[right]);
            if(height[left] > height[right]){
                res += rightMax - height[right];
                right--;
            }else{
                res += leftMax - height[left];
                left++;
            }
        }
        return res;
    }
}

43. 字符串相乘

题解

官方题解

思路

    m+n数组,第一个数中每一位数和第二个数中每一位数相乘,放到相应位置,然后进位得到结果

代码实现

class Solution {
    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0")) {
            return "0";
        }
        int m = num1.length(), n = num2.length();
        int[] ansArr = new int[m + n];
        for (int i = m - 1; i >= 0; i--) {
            int x = num1.charAt(i) - '0';
            for (int j = n - 1; j >= 0; j--) {
                int y = num2.charAt(j) - '0';
                ansArr[i + j + 1] += x * y;
            }
        }
        for (int i = m + n - 1; i > 0; i--) {
            ansArr[i - 1] += ansArr[i] / 10;
            ansArr[i] %= 10;
        }
        int index = ansArr[0] == 0 ? 1 : 0;
        StringBuffer ans = new StringBuffer();
        while (index < m + n) {
            ans.append(ansArr[index]);
            index++;
        }
        return ans.toString();
    }
}

44. 通配符匹配

题解

Linear runtime and constant space solution

i,j,startJ,lastMatch

class Solution {
    public boolean isMatch(String s, String p) {
        if(s == null || p == null){
            return false;
        }
        if(p.isEmpty()){
            return s.isEmpty();
        }
        int m = s.length(), n = p.length(), i = 0, j = 0, startJ = -1, lastMatch = -1;
        while(i < m){
            if(j < n && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?')){
                i++;
                j++;
            }else if(j < n && p.charAt(j) == '*'){
                startJ = j;
                j++;
                lastMatch = i;
            }else if(startJ != -1){
                j = startJ + 1;
                lastMatch++;
                i = lastMatch;
            }else{
                return false;
            }
        }
        while(j < n && p.charAt(j) == '*'){
            j++;
        }
        return j == n;
    }
}

45. 跳跃游戏 II

题解

官方题解

i == end 更新最远距离

class Solution {
    public int jump(int[] nums) {
        int n = nums.length, max = 0, end = 0, res = 0;
        for(int i = 0; i < n - 1; i++){
            max = Math.max(max, i + nums[i]);
            if(i == end){
                end = max;
                res++;
            }
        }
        return res;
    }
}
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